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Chapter 5: Problem 82

Eighty-six countries won medals at the 2016 Olympics in Rio de Janeiro. Based on the results: 1 country won more than 100 medals 2 countries won between 51 and 100 medals 3 countries won between 31 and 50 medals 4 countries won between 21 and 30 medals 15 countries won between 11 and 20 medals 15 countries won between 6 and 10 medals 46 countries won between 1 and 5 medals Suppose one of the 86 countries winning medals at the 2016 Olympics is selected at random. a. What is the probability that the selected country won more than 50 medals? b. What is the probability that the selected country did not win more than 100 medals? c. What is the probability that the selected country won 10 or fewer medals? d. What is the probability that the selected country won between 11 and 50 medals?

Short Answer

Expert verified
a. The probability that the selected country won more than 50 medals is \(\frac{3}{86}\). b. The probability that the selected country did not win more than 100 medals is \(\frac{85}{86}\). c. The probability that the selected country won 10 or fewer medals is \(\frac{61}{86}\). d. The probability that the selected country won between 11 and 50 medals is \(\frac{22}{86}\).

Step by step solution

01

Understanding probability and frequency distributions

Probability can be calculated as the ratio of the number of successful outcomes to the total number of possible outcomes. In this case, the possible outcomes are the 86 countries that won medals, and the successful outcomes depend on the conditions specified in each part of the question. A frequency distribution is a way to represent data by showing how often (frequency) different values of a variable occur. From the exercise, we already have the frequency distribution of medals won by the 86 countries:
02

Calculate the probabilities for part a

We're asked to find the probability that a randomly selected country won more than 50 medals. This occurs for 1 + 2 = 3 countries (1 country won more than 100 medals and 2 countries won between 51 and 100 medals). Hence, the successful outcome is 3. The probability is calculated as the ratio of successful outcomes over total outcomes, or in this case, \(\frac{3}{86}\). Therefore, the probability that the selected country won more than 50 medals is \(\frac{3}{86}\).
03

Calculate the probabilities for part b

We are asked to find the probability that the selected country did not win more than 100 medals. This scenario occurs for every country except the 1 country that won more than 100 medals. So, the successful outcome is 86 - 1 = 85. The probability is calculated as the ratio of successful outcomes over total outcomes, or in this case, \(\frac{85}{86}\). Therefore, the probability that the selected country did not win more than 100 medals is \(\frac{85}{86}\).
04

Calculate the probabilities for part c

We are asked to find the probability that the selected country won 10 or fewer medals. This occurs for 15 countries that won between 6 and 10 medals and 46 countries that won between 1 and 5 medals, a total of 15 + 46 = 61 countries. The probability is calculated as the ratio of successful outcomes over total outcomes, or in this case, \(\frac{61}{86}\). Therefore, the probability that the selected country won 10 or fewer medals is \(\frac{61}{86}\).
05

Calculate the probabilities for part d

We are asked to find the probability that the selected country won between 11 and 50 medals. This occurs for 3 countries that won between 31 and 50 medals, 4 countries that won between 21 and 30 medals, and 15 countries that won between 11 and 20 medals, a total of 3 + 4 + 15 = 22 countries. The probability is calculated as the ratio of successful outcomes over total outcomes, or in this case, \(\frac{22}{86}\). Therefore, the probability that the selected country won between 11 and 50 medals is \(\frac{22}{86}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Frequency Distribution
Frequency distribution is a fundamental concept in statistics that informs us about the spread of data points within a dataset. It provides a visual or tabular representation of how often each value occurs, which is critical in understanding the nature of the data at a glance.

For example, in an Olympic medal tally situation, we can create a frequency distribution table to show how many countries won medals within specific ranges鈥攍ike 1-5, 6-10, and so on. This tells us not only how many medals were won, but also how these medals are distributed across competing countries.

More formally, the frequency of an event is the number of times that event has occurred, and the distribution helps us organize these frequencies into an understanding of the data's pattern. By analyzing the frequency distribution, one can determine the most common outcomes, rare occurrences, or trends in the data, which is invaluable for making predictions or strategic decisions.
Calculating the Ratio of Successful Outcomes
The ratio of successful outcomes is the backbone of probability calculation. In a statistical context, a 'successful outcome' is defined as an outcome that meets the criteria we're interested in. For instance, when trying to determine the probability of selecting a country that won more than 50 medals from the list of medal winners, the successful outcomes would be the actual instances where this condition is met.

In the Olympic medal tally problem, there are 3 countries that won more than 50 medals, and hence, there are 3 successful outcomes. When we compare this to the total number of outcomes, which in this case is the total number of countries that won medals, we can see how these successful outcomes stack up against the whole. This ratio is the essence of calculating probabilities and offers a quantitative understanding of the likelihood of various events.
The Process of Probability Calculation
Probability calculation involves determining the chance of a specific outcome occurring within a set of possible outcomes. It's expressed as a value between 0 and 1, where 0 indicates impossibility, and 1 represents certainty. In the context of the Olympic medals example, calculating probabilities involves taking the ratio of successful outcomes to the total number of outcomes.

The formula for probability is conventionally understood as: \[ P(E) = \frac{{\text{{number of successful outcomes}}}}{{\text{{total number of outcomes}}}} \] where P(E) represents the probability of event E occurring.

To calculate these probabilities accurately, one must first identify the successful outcomes relevant to the question being asked, and then divide this number by the total number of possible outcomes. By applying this method to different scenarios in the Olympic medals example, we can derive probabilities for various conditions, such as not winning more than 100 medals or winning a swath of 11 to 50 medals. Understanding this process is essential for producing accurate statistical analyses in a wide range of fields.

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Most popular questions from this chapter

What does it mean to say that the probability that a coin toss will land head side up is \(0.5 ?\)

Airline tickets can be purchased online, by telephone, or by using a travel agent. Passengers who have a ticket sometimes don't show up for their flights. Suppose a person who purchased a ticket is selected at random. Consider the following events: \(O=\) event selected person purchased ticket online \(N=\) event selected person did not show up for flight Suppose \(P(O)=0.70, P(N)=0.07,\) and \(P(O \cap N)=0.04\) a. Are the events \(N\) and \(O\) independent? How can you tell? b. Construct a hypothetical 1000 table with columns corresponding to \(N\) and \(\operatorname{not} N\) and rows corresponding to \(O\) and \(\operatorname{not} O\). c. Use the table to find \(P(O \cup N)\). Give a relative frequency interpretation of this probability.

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A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

5.63 The report "Great Jobs, Great Lives. The Relationship Between Student Debt, Experiences and Perceptions of College Worth" (Gallup-Purdue Index 2015 Report) gave information on the percentage of recent college graduates (those graduating between 2006 and \(2015,\) inclusive \()\) who strongly agree with the statement "My college education was worth the cost." Suppose that a college graduate will be selected at random, and consider the following events: \(A=\) event that the selected graduate strongly agrees that education was worth the cost \(N=\) event that the selected graduate finished college with no student debt \(H=\) event that the selected graduate finished college with high student debt (over \(\$ 50,000)\) The following probability estimates were given in the report: $$ P(A)=0.38 \quad P(A \mid N)=0.49 \quad P(A \mid H)=0.18 $$ a. Interpret the value of \(P(A \mid N)\). b. Interpret the value of \(P(A \mid H)\). c. Are the events \(A\) and \(H\) independent? Justify your answer.
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